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Theorem elabg 2682
 Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)
Hypothesis
Ref Expression
elabg.1 (x = A → (φψ))
Assertion
Ref Expression
elabg (A 𝑉 → (A {xφ} ↔ ψ))
Distinct variable groups:   ψ,x   x,A
Allowed substitution hints:   φ(x)   𝑉(x)

Proof of Theorem elabg
StepHypRef Expression
1 nfcv 2175 . 2 xA
2 nfv 1418 . 2 xψ
3 elabg.1 . 2 (x = A → (φψ))
41, 2, 3elabgf 2679 1 (A 𝑉 → (A {xφ} ↔ ψ))
 Colors of variables: wff set class Syntax hints:   → wi 4   ↔ wb 98   = wceq 1242   ∈ wcel 1390  {cab 2023 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553 This theorem is referenced by:  elab2g  2683  intmin3  3633  finds  4266  elxpi  4304  ovelrn  5591  indpi  6326  peano5nni  7678  peano5set  9374
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